# When is it okay to use $Q=mc\Delta T$, is this equation only for calorimetry questions?

When is it okay to use $Q=mc\Delta T$, is the equation only for calorimetry questions? When a question talks about heat how do I know if i'm measuring $\Delta H$ or Q?

The equation $Q = mc \Delta T$ is appropriate for calorimetry at a constant volume. $Q$ here refers to the net heat transfer of the isolated system. Change in enthalpy, $\Delta H$, on the other hand, is really the total change in energy of the system, that is:

$$\Delta H = \Delta U + P\Delta V + V\Delta P$$

Where $P$ is pressure, $V$ is volume, and $U$ is the internal energy of the system, which is the sum of the system's net heat transfer ($Q$) and the work done on the system ($W$): $\Delta U = Q + W$. If we substitute the definition of $\Delta U$ into our equation for enthalpy, we get:

$$\Delta H = Q + W + P\Delta V + V\Delta P$$

Because any mechanical pressure-volume work performed on a closed system is equal to $-P\Delta V$, the equation further reduces to:

$$\Delta H = Q + V\Delta P$$

From this it becomes apparent that $Q$ and $\Delta H$ are only equal if pressure is constant (i.e., $\Delta P = 0$). The upshot of all this is that you should not assume $\Delta H$ and $Q$ are equal unless you're explicitly instructed to assume constant volume and pressure. When a problem refers to heat, it's almost certainly referring to heat transfer, $Q$, and not enthalpy change ($\Delta H$).

• There needs to be a $V \Delta P$ term somewhere, otherwise, why is constant pressure important? Work is ($w=-P\Delta V$), so your second equation simplifies to $\Delta H = q$ even if the volume changes. At constant volume $q_v=\Delta U$ ($w=0$) and you don't have to invoke enthalpy at all. Commented Apr 18, 2013 at 10:25

You can use $q=mc\Delta T$ for any process in which heat is transferred, not just calorimetry.

Generally, when a question talks about heat transfer, that means $q$. If the question meant enthalpy change, the question would have used the word "enthalpy".

Under constant pressure conditions, the enthalpy change is equal to the heat transfer. Enthalpy equals internal energy plus $PV$. Thus, change in enthalpy is:

$$H=U+PV$$ $$\Delta H = \Delta U + P\Delta V + V\Delta P$$

Since the change in internal energy is equal to the sum of heat transfer and work (and work is $w=-P\Delta V$):

$$\Delta H = q -P\Delta V + P\Delta V + V\Delta P = q+ V\Delta P$$

At constant pressure, $q=q_p$ and $\Delta P = 0$, so

$$\Delta H = q_p = mc_p \Delta T$$

Constant pressure is common in the lab when the reaction vessel is open to the air. At constant pressure, if you are measuring $q$, you are also measuring $\Delta H$.